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IrinaK [193]
3 years ago
12

What will happen to a test dummy during a crash test if its seatbelt is not buckled? The test dummy will stop when the car stops

. The test dummy will continue forward until it makes contact with another object. The test dummy will strike and rebound continuously within the interior of the car The test dummy will remain motionless.
Chemistry
2 answers:
Arada [10]3 years ago
8 0

Answer:

The test dummy will continue forward until it makes contact with another object.

Explanation:

When a car brakes abruptly the body of the test dummy is thrown forward, the law of inertia comes into play. Therefore, the use of the seat belt is essential to ensure the integrity of the passengers.

The test dummy is at rest relative to the car and in relation to the ground, the speed of the body of that doll is zero. A collision of the car with some object pushes the puppet forward, causing it to vary from zero speed to a nonzero speed.

If a test dummy is buckled, but another test dummy in the back seat of the car is without the belt, at the time of braking, it will be thrown in relation to the car at the speed that the car had instantly before the moment of braking. That is, test dummy will continue to advance until it contacts another object.

algol133 years ago
4 0
<span>The test dummy will continue forward until it makes contact with another object.</span>
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A 85.2 g copper bar was heated to 221.32 degrees Celsius and placed in a coffee cup calorimeter containing 425 mL of water at 22.55 degrees Celsius. The final temperature of the water was recorded to be 26.15 degrees Celsius. What is the specific heat of the copper?

Step 1: Data given

Mass of copper = 85.2 grams

Temperature of copper = 221.32 °C

Volume of water = 425 mL

Temperature of water = 22.55 °C

Final temperature = 26.15 °C

Specific heat of water = 4.184 J/g°C

Step 2: Calculat the specific heat of copper

Heat lost = heat gained

Q = m*c*ΔT

Qcopper = -Qwater

m(copper)*c(copper)*ΔT(copper) = - m(water) * c(water) * ΔT(water)

⇒ m(copper) = 85.2 grams

⇒ c(copper) = TO BE DETERMINED

⇒ ΔT(copper) = the change in temeprature = T2 -T1 = 26.15 -221.32 = -195.17 °C

⇒ m(water) = The mass of water = 425 mL * 1g/mL = 425 grams

⇒ c(water) = The specific heat of water = 4.184 J/g°C

⇒ ΔT(water) = The change of temperature of water = 26.15 - 22.55 = 3.6

85.2 * c(copper) * (-195.17) = -425 * 4.184 * 3.6

c(copper) = 0.385 J/g°C

The specific heat of copper is 0.385 J/g°C

(Note, The original question says the volume of the water is 4250 mL. IF this is not an error, the specific heat of copper is 3.85 J/g°C (10x higher than the normal value).

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